The following question comes from Some integral with sine post
$$\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$$
but now I'd be curious to know how to deal with it by methods of complex analysis.
Some suggestions, hints? Thanks!!!

I think someone made a paper about this, but can't recall a source.
–
Pedro Tamaroff♦Feb 18 '13 at 22:37

5

@PeterTamaroff You must be referring to the "Surprising sinc sums and integrals" by Baillie, Borwein and Borwein. ( pdf )
–
SashaFeb 18 '13 at 22:50

4

Note that the integral has an interpretation of $2 f_{X_n}(0)$, where $X_n$ is the sum of $n$ uniform on $(-1,1)$ random variables, and $f_X(x)$ denotes the pdf. Using the central limit theorem, one can find large $n$ asymptotics, $\int_0^\infty \sin^n(x)/x^n \mathrm{d} x \approx \sqrt{\frac{3 \pi}{2 n}} $
–
SashaFeb 18 '13 at 23:00

+1 for reminding me of the Irwin-Hall distribution. The sum you derives is exactly the $2 f_X(0)$, where $X$ is the Irwin-Hall random variable.
–
SashaFeb 19 '13 at 1:09

@Sasha: Indeed it appears that the integral is $\frac{\pi}{2}f_X(\frac{n}{2},n)$. Thank you for informing me of this interesting connection!
–
user26872Feb 19 '13 at 2:11

@oen: good shot! Long time I haven't seen you around! So, welcome! :-) (+1)
–
Chris's sis the artistFeb 19 '13 at 9:33

@Chris'ssisterandpals: Glad to help. Your questions often have connections to more areas of mathematics than it appears at first glance, which make them very interesting indeed.
–
user26872Feb 19 '13 at 11:58

Just to verify oen's post (since there is a post with a different answer), I will post the answer I got.

$|\sin(z)|\le e^{|\mathrm{Im}(z)|}$; therefore, on the strip $|\mathrm{Im}(z)|\le1$, we have $|\sin(z)|\le e$. Thus, $\left(\frac{\sin(z)}{z}\right)^n$ vanishes as $|z|\to\infty$ in that strip and therefore,
$$
\int_{-\infty}^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z
=\int_{-\infty-i}^{\infty-i}\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z\tag{1}
$$
Next define two contours $\gamma^+$ and $\gamma^-$. $\gamma^+$ goes from $-R-i$ to $R-i$ then circles back through the upper half plane along $|z+i|=R$. $\gamma^-$ goes from $-R-i$ to $R-i$ then circles back through the lower half plane along $|z+i|=R$.

Using the binomial theorem, we get
$$
\left(\frac{\sin(z)}{z}\right)^n=\frac1{(2iz)^n}\sum_{k=0}^n(-1)^k\binom{n}{k}e^{(n-2k)iz}\tag{2}
$$
Integrate the terms where $n-2k\ge0$ along $\gamma^+$ and the others along $\gamma^-$. Since $\gamma^-$ doesn't enclose any singularities, we can ignore that integral. Therefore,
$$
\begin{align}
\int_0^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z
&=\frac12\int_{\gamma^+}\frac1{(2iz)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}e^{(n-2k)iz}\mathrm{d}z\\
&=\frac{\pi i}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\mathrm{Res}\left(\frac{e^{(n-2k)iz}}{z^n},0\right)\\
&=\frac{\pi i}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\frac{(n-2k)^{n-1}i^{n-1}}{(n-1)!}\\
&=\frac{\pi}{2^n(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}(n-2k)^{n-1}\tag{3}
\end{align}
$$

First, to simplify matters let's take $n$ odd and $\geq 3$. Let $C_{\epsilon}^+$ be the contour along the real line that takes a semicircular detour into the upper half plane about the origin, and let $C_{\epsilon}^-$ be the same for the lower half plane. We use continuity of the integrand to argue that
$$
I = \lim_{\epsilon \rightarrow 0} \int_{C_{\epsilon}^{\pm}} = \frac{1}{2} \lim_{\epsilon \rightarrow 0} \left( \int_{C_{\epsilon}^+} + \int_{C_{\epsilon}^-} \right)
$$
Now think about $(\sin x)^n$: it's a sum of exponential terms of the form $e^{i l x}$ for $-n \leq l \leq n$ with some coefficients. You should convince yourself that any $l < 0$ term is killed by $\int_{C_{\epsilon}^-}$ and any $l > 0$ term is killed by $\int_{C_{\epsilon}^+}$. Moreover by completing these contours with large semicircles, you can derive ($l > 0$):
$$
\int_{C_{\epsilon}^{\mp}} \frac{e^{\pm i l x}}{x^n} dx = \mp 2 \pi i \frac{(\pm i l)^{n-1}}{(n-1)!}
$$
Summing everything up and noticing that there is no $\epsilon$ dependence, and keeping track of signs (which I failed to do on a first pass) we've shown that,
$$
I = \frac{\pi }{2^{n-1} (n-1)!} \sum_{l = 0}^{(n-1)/2} (-1)^{n-1-l}\left(\begin{array}{c}n \\ l \end{array} \right) (n-2l)^{n-1}
$$
I hope that wasn't too much (or too little).

You've written $I$ to be twice the asked integral. However, when $n=2$, your answer gives $-\pi$ as the integral of a positive function. The problem seems to be that the exponent of $-1$ has an extraneous $n-1$.
–
robjohn♦Feb 19 '13 at 11:05

@oen: Ah, I hadn't noticed that this answer was restricted to odd $n$. That assumption doesn't seem to be used in any part of the proof. Thanks for the upvote; I made my answer CW in hopes that it would not steal any votes.
–
robjohn♦Feb 19 '13 at 12:14

@robjohn: That is very considerate, but your answer deserves upvotes!
–
user26872Feb 19 '13 at 12:27

@oen: I just looked more closely at your answer and I see it is a different approach from mine. You changed the integrand where I changed the contour. Perhaps I will remove the CW :-)
–
robjohn♦Feb 19 '13 at 12:37